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Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 = 25
25 is the perfect square of 5. rest all options are not perfect squares of any natural number.
LCM = 2×3×5×2×3×5×4 = 3600
Factors of = 25 = 5^{2} are 1, 5, 25 which can also be represented as: 5^{2 }= (2+1) = 3 (i.e.3 factors)
It is divisible by 8 & 3 because 8 and 3are coprime
512×343 =
2×2×2×2×2×2×2×2×2×7×7×7=2^{9} x 7^{3}
Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21?
Reminder= 40 84040=800 100000+800=100800
HCF of 81 and 135 = 27 HCF of 87 and 116 = 29 HCF of 43 and 73 = 1 HCF of 26 and 65 = 13 Hence, 43 and 73 are coprime
1024=2^{10}=2^{n}
⇒ (n+1) = (10+1) = 11.
Where (n+1) = no. of factors. Here factors will be 1, 2, 4, 8, 16?????.1024.
If the 8digit number 136y5785 is divisible by 15, then find the least possible value of
136Y5785
For 15, No. should be divisible by 5, 3
No. is already divisible by 5 1 + 3 + 6 + 5 + 7 + 8 + 5 + y = 35 + y
∴y=1
Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 in each case?
5×4×5×2×3=600
Required numbers is 600+7=607
x, y, z, w are four odd natural numbers. Let u=x^{2}+y^{2}+z^{2}+w^{2}. Consider following statements
I. U is always divisible by 4
II. U is never divisible by 8
Q. Which of the above statements) is /are true?
2211/5025 , which reduced to its simplest form. By dividing the numerator and denominator
A boy saves Rs. 4.65 daily. Find the least number of days in which he will be to save an exact number (whole number) of rupees.
Rs. 4.65 = 465 paise
The exact number of rupees will be a multiple of 100
So, we have to fine the L.C.M. of 465 and 100
100 = 2*2*5*5
465 = 3*5*31
L.C.M. of 465 and 100 = 2*2*3*5*5*31
= 9300
So, the boy saves 9300 paise
Hence, least number of days required to save 9300 paise = 9300/465
= 20 days
H.C.F is always a factor of L.C.M. In this case 16 Divides 192 only.
29 videos73 docs45 tests

29 videos73 docs45 tests
